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3 years ago

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If you started the motor of a boat in the middle of a lake, who would detect the sound of the motor first: a friend sitting on t

he shore of the lake, or a friend snorkeling just below the surface of the water along the same shore? Explain your answer.

2 answers:

Andrei [34K]3 years ago

7 0

Answer:

A friend snorkeling just below the surface of the water along the same shore will detect the sound first.

ArbitrLikvidat [17]3 years ago

6 0

Answer:

A friend snorkeling just below the surface of the water along the same shore will detect the sound first.

Explanation:

The speed of sound in water medium is faster than that through the air.
Sound propagates through the medium by transferring through the molecules on it. Water has more closely packed molecules due to which the speed is faster.
In fact, the sound's speed in water is almost four times faster than that in the air.
So the guy in the water surface gets to hear sound faster than the one in sore.

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3 years ago

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3 years ago

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Two satellites, X and Y, are orbiting Earth. Satellite X is 1.2 × 106 m from Earth, and Satellite Y is 1.9 × 105 m from Earth. W

jenyasd209 [6]

Answer: Satellite X has a greater period and a slower tangential speed than Satellite Y

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

$T^{2}=\frac{4\pi^{2}}{GM}r^{3}$ (1)

Where;

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24}kg$ is the mass of the Earth

$r$ is the semimajor axis of the orbit each satellite describes around Earth (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

So for satellite X, the orbital period $T_{X}$ is:

$T_{X}^{2}=\frac{4\pi^{2}}{GM}r_{X}^{3}$ (2)

Where $r_{X}=1.2(10)^{6}m$

$T_{X}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.2(10)^{6}m)^{3}$ (3)

$T_{X}=413.712 s$ (4)

For satellite Y, the orbital period $T_{Y}$ is:

$T_{Y}^{2}=\frac{4\pi^{2}}{GM}r_{Y}^{3}$ (5)

Where $r_{Y}=1.9(10)^{5}m$

$T_{Y}^{2}=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}(1.9(10)^{5}m)^{3}$ (6)

$T_{Y}=26.064 s$ (7)

This means $T_{X}>T_{Y}$

Now let's calculate the tangential speed for both satellites:

<u>For Satellite X:</u>

$V_{X}=\sqrt{\frac{GM}{r_{X}}}$ (8)

$V_{X}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.2(10)^{6}m}}$

$V_{X}=18224.783 m/s$ (9)

<u>For Satellite Y:</u>

$V_{Y}=\sqrt{\frac{GM}{r_{Y}}}$ (10)

$V_{Y}=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24}kg)}{1.9(10)^{5}m}}$

$V_{Y}= 45801.13 m/s$ (11)

This means $V_{Y}>V_{X}$

Therefore:

Satellite X has a greater period and a slower tangential speed than Satellite Y

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3 years ago

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